Implicitly Defined Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.8: “Implicit Differentiation” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First
Look at the equation $x^2 + y^2 = 25$.
Predict: does this describe $y$ as a function of $x$? Sketch the curve quickly. How many $y$-values correspond to $x = 3$?
If you have two possible $y$-values at the same $x$, then $y$ is not a function of $x$ -- at least not globally. But near the point $(3, 4)$, $y$ behaves like a function of $x$: as $x$ increases slightly from 3, $y$ decreases slightly from 4 along the upper arc.
An “implicitly defined function” is a function whose values come from an equation in $x$ and $y$ rather than from a formula $y = f(x)$. The next sections make this precise.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
Explicit function: $y = f(x)$ -- $y$ is written directly in terms of $x$.
Implicit function: $F(x, y) = 0$ -- $y$ and $x$ appear together in an equation; $y$ is not isolated.
Local implicit function: Near a specific point $(a, b)$ on the curve $F(x,y) = 0$, $y$ may be treatable as a smooth function of $x$ even if it is not globally.
Key Concepts
1. Explicit vs. Implicit
Explicit: $y = 3x^2 - 5x + 1$. The variable $y$ is isolated on the left.
Implicit: $x^2 + y^2 = 25$. Both $x$ and $y$ appear on the same side; $y$ is not isolated.
Some implicit equations can be solved for $y$ explicitly; many cannot. The circle $x^2 + y^2 = 25$ gives $y = \pm\sqrt{25 - x^2}$ -- two branches, not a single function. The curve $x^3 + y^3 = 6xy$ (the folium of Descartes) cannot be solved for $y$ in a simple closed form.
2. Multiple Representations: Points, Graphs, and Equations
Equation form: $x^2 + y^2 = 25$. Tells you the rule for which $(x,y)$ pairs lie on the curve.
Graphical form: A circle of radius 5 centered at the origin. The curve fails the vertical line test globally (each $x \in (-5, 5)$ corresponds to two $y$-values), so $y$ is not a global function of $x$.
Local function: Near $(3, 4)$: $y = \sqrt{25 - x^2}$ (upper branch). Near $(3, -4)$: $y = -\sqrt{25 - x^2}$ (lower branch). Locally, $y$ is a function of $x$ on each branch.
Table of points on $x^2 + y^2 = 25$:
| $x$ | $y$ (upper branch) | $y$ (lower branch) |
|---|---|---|
| $0$ | $5$ | $-5$ |
| $3$ | $4$ | $-4$ |
| $4$ | $3$ | $-3$ |
| $5$ | $0$ | $0$ |
3. Why Implicit Functions Arise
Many curves in mathematics and applications are defined implicitly:
- Conic sections: Ellipses ($\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$), hyperbolas ($\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$).
- Level curves of surfaces: In thermodynamics, $pV = nRT$ relates pressure, volume, and temperature; solving for $V$ explicitly requires isolating one variable.
- Algebraic curves: $y^3 - xy = 1$ defines $y$ implicitly and cannot be solved in closed form.
Implicit differentiation lets you find the slope of these curves without isolating $y$.
4. Treating $y$ as a Function of $x$
The key assumption for implicit differentiation: even when $y$ is not explicitly isolated, we treat $y$ as a differentiable function of $x$ near the point of interest. This means every time we differentiate a term involving $y$ with respect to $x$, the chain rule applies: $y$ is the “inner function.”
Example 1. If $y$ is a function of $x$, what is $\dfrac{d}{dx}[y^2]$?
By the chain rule (outer $= t^2$, inner $= y(x)$): \[ \frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx}. \]
Example 2. What is $\dfrac{d}{dx}[\sin y]$? \[ \frac{d}{dx}[\sin y] = \cos y \cdot \frac{dy}{dx}. \]
Example 3. What is $\dfrac{d}{dx}[xy]$?
Product rule (both $x$ and $y$ are functions of $x$): \[ \frac{d}{dx}[xy] = y + x\frac{dy}{dx}. \]
5. Ask Why: When Is the Implicit Function Theorem Needed?
At a point $(a, b)$ on $F(x, y) = 0$, the curve locally defines $y$ as a function of $x$ precisely when the partial derivative $\partial F / \partial y \neq 0$ at that point. This is the Implicit Function Theorem (a calculus II or analysis result).
For MATH161, you will typically be told to assume $y$ is a differentiable function of $x$ near the point of interest. Understanding that this assumption can fail -- at points where the curve has a vertical tangent or self-intersects -- explains why you should always state “near $(a, b)$” when using implicit differentiation.
Named Misconception: “Implicit Functions Are Not Functions”
A curve defined by $F(x, y) = 0$ may not pass the vertical line test globally, so it is not a single function on its entire domain. But locally, near a point where the curve is smooth, it does behave as a function. “Implicit function” means a function defined implicitly by a relation, not that it fails to be a function.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Forgetting the chain rule | $\frac{d}{dx}[y^2] = 2y$ | Must include $dy/dx$: $\frac{d}{dx}[y^2] = 2y\frac{dy}{dx}$ |
| Treating $y^2$ as if $y$ were constant | Differentiating $y^2$ as $0$ | $y$ depends on $x$; the chain rule gives $2y\frac{dy}{dx}$ |
Common Misconceptions
treating $y$ as a constant when differentiating an implicit equation with respect to $x$.
This is the composition-is-not-chaining error. If $y$ is treated as a constant, then $\dfrac{d}{dx}[y^2]$ yields $0$, and $\dfrac{d}{dx}[xy]$ yields just $y$. Both are wrong. Because $y$ is a function of $x$ along the curve, the chain rule applies: $\dfrac{d}{dx}[y^2] = 2y\dfrac{dy}{dx}$ and $\dfrac{d}{dx}[xy] = y + x\dfrac{dy}{dx}$. Treating $y$ as a constant produces a result that ignores how $y$ tracks with $x$ along the curve.
Leveled Practice
Level 1 -- Differentiating $y$-Terms
Problem 1. Compute $\dfrac{d}{dx}$ of each expression, treating $y$ as a function of $x$.
(a) $y^3$, (b) $\cos y$, (c) $y^4 + 2y$
Show answer
(a) $3y^2 \frac{dy}{dx}$.
(b) $-\sin y \frac{dy}{dx}$.
(c) $(4y^3 + 2)\frac{dy}{dx}$.
Problem 2. Compute $\dfrac{d}{dx}[x^2 y]$.
Show answer
Product rule: $2xy + x^2\frac{dy}{dx}$.
Level 2 -- Identifying Explicit Branches
Problem 3. For the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$:
(a) Solve for $y$ explicitly and identify the two branches.
(b) Evaluate both branches at $x = 0$ and $x = 3$.
Show answer
(a) $y = \pm 2\sqrt{1 - x^2/9} = \pm \frac{2}{3}\sqrt{9 - x^2}$.
(b) At $x = 0$: $y = \pm 2$. At $x = 3$: $y = 0$ (the curve meets the $x$-axis; only one branch here).
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 4 (Extension).
(a) (Floor) For $x^2 + y^2 = 1$: at which point(s) does the curve NOT locally define $y$ as a function of $x$? Explain geometrically.
(b) (Mid) Write $F(x,y) = x^2 + y^2 - 1$. Compute $\partial F / \partial y$ and verify it is zero exactly at the points you found in (a).
(c) (Ceiling) The folium of Descartes is $x^3 + y^3 = 3xy$. Verify that $(3/2, 3/2)$ lies on the curve. At this point, does the curve locally define $y$ as a function of $x$? Why might you need to check?
Show answer
(a) At $(\pm 1, 0)$, the circle has a vertical tangent (the curve crosses the vertical line $x = a$ at two nearby points). Near $(\pm 1, 0)$, $y$ is NOT a function of $x$.
(b) $\partial F/\partial y = 2y$. At $y = 0$ (i.e., $(\pm 1, 0)$): $2y = 0$. The partial derivative is zero exactly at the problematic points.
(c) $(3/2)^3 + (3/2)^3 = 2(27/8) = 27/4$; $3 \cdot (3/2)(3/2) = 3 \cdot 9/4 = 27/4$. Verified.
At $(3/2, 3/2)$: $\partial F/\partial y = 3y^2 - 3x = 3(9/4) - 3(3/2) = 27/4 - 9/2 = 9/4 \neq 0$. So $y$ is locally a function of $x$ at this point (the Implicit Function Theorem applies).
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
An implicit equation is a filter: it picks out all $(x, y)$ pairs that satisfy the relation. The curve it defines may not pass the vertical line test globally -- you cannot read a single $y$ from each $x$. But near any smooth, non-vertical point on the curve, $y$ tracks $x$ in a well-defined, differentiable way.
Implicit differentiation exploits this: even without solving for $y$, you differentiate both sides of the equation with respect to $x$, treating $y$ as an unknown function of $x$ and applying the chain rule wherever $y$ appears.
Connections
Within Calculus I (MATH161)
- Implicit differentiation: The key idea here, treating $y$ as a function of $x$, drives the full procedure in the next lesson.
- Related rates: Related-rate problems use the same “treat both $x$ and $y$ as functions of $t$” idea. The equation relating $x$, $y$, and possibly other variables is differentiated with respect to $t$.